CBSE Notes Class 10 Maths Chapter 7 Coordinate Geometry

CBSE Notes Class 10 Maths Chapter 7 – Coordinate Geometry offer a simple and structured understanding of how algebra connects with geometry through the coordinate plane. In this chapter, students learn important concepts such as the distance formula, section formula, and how to find the area of a triangle using coordinates. These CBSE Notes present formulas, explanations, and solved examples in a clear format, helping students revise quickly and strengthen their problem-solving skills for the board exams.

1.0Download CBSE Notes for Class 10 Maths Chapter 7: Coordinate Geometry - Free PDF!!

Download the CBSE Notes for Class 10 Maths Chapter 7: Coordinate Geometry in free PDF format for quick revision. Access clear formulas, key concepts, and solved examples designed for effective board exam preparation.

2.0CBSE Class 10 Maths Chapter 7 Coordinate Geometry - Revision Notes

What is the Cartesian Coordinate System?

1. It is a two-dimensional Cartesian coordinate system in which the position of any point can be defined with the aid of two perpendicular lines, and these lines are called axes.

• The horizontal axis is named as the x-axis or Abscissa.
• The y-axis is the vertical axis or Ordinate. 

2. If the x-axis and the y-axis intersect at a point, such an intersection is called origin. The origin is denoted by the letter O with the coordinate (0, 0).
3. Any point in the coordinate plane is represented as P(x, y), where x denotes the distance from the origin along the x-axis and y is the distance from the origin along the y-axis.

The Distance Formula 

In maths, the distance formula is used to find the distance between two points on the cartesian system, for example A(x₁,y₁) and B(x₂,y₂). The distance is always positive. The formula to find the distance between these points: 

$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance is two points if any one point is the origin.  $AB=\sqrt{(x{)}^2+(y{)}^2}$

Example 1: Find the distance between P(2, 6) and Q(8, 3) using the distance formula. 

Solution:

$PQ=\sqrt{(8-2{)}^2+(3-6{)}^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$

Example 2: Find the point on the y-axis which is equidistant from A(2, 4) and B(4, 8). 

Solution: The point is on the y-axis, hence the coordinates will be (0, y)

$\sqrt{(0-2{)}^2+(y-4{)}^2}=\sqrt{(0-4{)}^2+(y-8{)}^2}$

Squaring both sides,

4 + y² - 8y + 16 = 16 + y² - 16y + 64 

-8y + 16y = 64 - 4

8y = 60 

y = 7.5

Note: If the given point is on the x-axis, then the y-coordinate will be 0 and vice-versa. 

Section Formula 

Section formula is used to find the coordinates of a line segment. For example, in the given figure, C(x, y) is the point cutting line segments A(x₁, y₁) and B(x₂, y₂) in the ratio m and n.

$\mathbf{X}=\frac{m\times x_2+n\times x_1}{m+n}$

$\mathrm{y}=\frac{m\times y_2+n\times y_1}{m+n}$

Here, (x, y) are the coordinates of C. 

Midpoint Formula: 

When point C cuts the line segment AB in two equal parts, meaning m = n, then we can use another formula that is:

$\mathrm{X}=\frac{x_2+x_1}{2}$

$\mathrm{y}=\frac{y_2+y_1}{2}$

Example: Find the coordinates of a point R(x, y) which is cutting the line segment P(2, 4) and Q(4, 8) in the ratio of 2 and 4. 

Solution:

$X=\frac{2\times 4+4\times 2}{2+4}=\frac{8+8}{6}=\frac{8}{3}$

$y=\frac{2\times 8+4\times 4}{2+4}=\frac{16+16}{6}=\frac{16}{3}$

$\text{ Coordinates of }\mathrm{P}\left(\frac{8}{3},\frac{16}{3}\right)\text{. }$

Example: Find the ratio of the point Q that is on the y-axis cutting the line segment R(5, 7) and P(9, 3). 

Solution: Let the ratio be K : 1.

The point Q is on the y-axis; hence, Q(0, y)

Coordinates of Q 

$0=\frac{k\times 9+1\times 5}{K+1}$

k9+15 = 0 

K = 9/5

Hence, The ratio is 9 : 5 

3.0CBSE Class 10 Maths Chapter 7 Coordinate Geometry - Key Notes

Cartesian plane

The horizontal line is called the x-axis and the vertical line is called y-axis.
The point where these lines intersect is called the origin. Cartesian plane is also known as x, y plane.

I quadrant → x > 0, y > 0
II quadrant → x < 0, y > 0
III quadrant → x < 0, y < 0
IV quadrant → x > 0, y < 0

Distance formula

Distance between two points P(x₁, y₁) and Q(x₂, y₂) is given by

$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

Distance of a point from origin

The distance of a point (x, y) from origin is $\sqrt{x^2+y^2}$

Test for Geometrical Figures

For an isosceles triangle : Prove that two sides are equal.

For an equilateral triangle : Prove that threes sides are equal.

For a right-angled triangle : Prove that the sum of the squares of two sides is equal to the square of the third side.

For a square : Prove that all sides are equal and diagonals are equal.

For a rhombus : Prove that all sides are equal and diagonals are not equal.

For a rectangle : Prove that the opposite sides are equal and diagonals are also equal.

For a parallelogram : Prove that the opposite sides are equal in length and diagonals are not equal.

Collinearity of three points

Let A, B and C are three given points. Point A, B and C will be collinear, if the sum of lengths of any two line-segments is equal to the length of the third line-segment.

(i) AB + BC = AC
(ii) AB + AC = BC
(iii) AC + BC = AB

Section formula

Formula for internal division

The coordinates of the point R(x, y) which divides the join of two given points P(x₁, y₁) and Q(x₂, y₂) internally in the ratio m₁ : m₂ are given by $\left(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2}\right)$

Formula for external division

The coordinates of the point R(x, y) which divides the join of two given points P(x₁, y₁) and Q(x₂, y₂) externally in the ratio m₁ : m₂ are given by$\left(\frac{m_1{x}_2-m_2{x}_1}{m_1-m_2},\frac{m_1{y}_2-m_2{y}_1}{m_1-m_2}\right)$

Mid-point formula

If R is the midpoint of PQ, then its coordinates are given by$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Centroid of a triangle

The point of concurrence of the medians of a triangle is called the centroid of triangle. It divides the median in the ratio 2 : 1.

The coordinates of the centroid of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) are given by$\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$

Area of Triangle

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by

$A=\frac{1}{2}\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$

Condition of collinearity of three points

The points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) will be collinear (i.e., will lie on a straight line) if the area of the triangle assumed to be formed by joining them is zero.

4.0Key Features of CBSE Maths Notes for Class 10 Chapter 7

• The notes are well aligned with the latest curriculum suggested by CBSE. 
• Every concept is simplified and explained with simple and easy language, making these notes ideal for self-learning. 
• The step-by-step guide is provided with every concept of coordinate geometry to ensure better understanding.